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Question: A chemical reaction \(2A\to 4B+C\); in a gaseous phase shows an increase in concentration of B by \(...

A chemical reaction 2A→4B+C2A\to 4B+C; in a gaseous phase shows an increase in concentration of B by 5×10−3M5\times{{10}^{-3}}Min 10 second. Calculate:
(a) Rate of appearance of B.
(b) Rate of the reaction.
(c) Rate of disappearance of A.

Explanation

Solution

Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time.

Complete step by step answer:
Rate is defined as the change in concentration of reactant or product with respect to time. Let us first understand what the term rate of appearance and rate of disappearance means.
The rate of a reaction is observed by watching the rate of disappearance of reactant and the appearance of product over a unit time.
Given reaction is 2A→4B+C2A\to 4B+C.
First part:
Given increase in concentration of B = 5×10−3M5\times{{10}^{-3}}M
Rate of appearance of B will be the ratio of increase in concentration of B which is produced in a unit time.
Rate = 5×10−3(molL−1)10(sec⁡)=5×10−4molL−1s−1\dfrac{5\times{{10}^{-3}}(mol{{L}^{-1}}) }{10(\sec ) }=5\times{{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}
Rate of appearance of B = 5×10−4molL−1s−15\times{{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}
Second part:
Rate of the reaction = −12d[A]dt=14d[B]dt-\dfrac{1}{2}\dfrac{d[A]}{dt}=\dfrac{1}{4}\dfrac{d[B]}{dt}
−12d[A]dt-\dfrac{1}{2}\dfrac{d[A]}{dt} = rate of disappearance of A
14d[B]dt\dfrac{1}{4}\dfrac{d[B]}{dt} = rate of appearance of B
Rate of reaction = 14d[B]dt=14×5×10−4\dfrac{1}{4}\dfrac{d[B]}{dt}=\dfrac{1}{4}\times5\times{{10}^{-4}}
=1.25×10−4molL−1s−1=1.25\times{{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}
Third part:
Rate of disappearance of A = −12d[A]dt-\dfrac{1}{2}\dfrac{d[A]}{dt}
Rate of reaction is 1.25×10−4molL−1s−11.25\times{{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}
Rate of disappearance of A = 2 \times 1.25×10−4molL−1s−11.25\times{{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}
= 2.5×10−4molL−1s−12.5\times{{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}

Note: Rate of the reaction = −12d[A]dt=14d[B]dt-\dfrac{1}{2}\dfrac{d[A]}{dt}=\dfrac{1}{4}\dfrac{d[B]}{dt}. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing.